Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $t = \dfrac{x - 9}{-3x + 30} \times \dfrac{5x^3 - 100x^2 + 500x}{x^3 - 3x^2 - 54x} $
Explanation: First factor out any common factors. $t = \dfrac{x - 9}{-3(x - 10)} \times \dfrac{5x(x^2 - 20x + 100)}{x(x^2 - 3x - 54)} $ Then factor the quadratic expressions. $t = \dfrac {x - 9} {-3(x - 10)} \times \dfrac {5x(x - 10)(x - 10)} {x(x - 9)(x + 6)} $ Then multiply the two numerators and multiply the two denominators. $t = \dfrac {(x - 9) \times 5x(x - 10)(x - 10) } {-3(x - 10) \times x(x - 9)(x + 6) } $ $t = \dfrac {5x(x - 10)(x - 10)(x - 9)} {-3x(x - 9)(x + 6)(x - 10)} $ Notice that $(x - 9)$ and $(x - 10)$ appear in both the numerator and denominator so we can cancel them. $t = \dfrac {5x(x - 10)(x - 10)\cancel{(x - 9)}} {-3x\cancel{(x - 9)}(x + 6)(x - 10)} $ We are dividing by $x - 9$ , so $x - 9 \neq 0$ Therefore, $x \neq 9$ $t = \dfrac {5x\cancel{(x - 10)}(x - 10)\cancel{(x - 9)}} {-3x\cancel{(x - 9)}(x + 6)\cancel{(x - 10)}} $ We are dividing by $x - 10$ , so $x - 10 \neq 0$ Therefore, $x \neq 10$ $t = \dfrac {5x(x - 10)} {-3x(x + 6)} $ $ t = \dfrac{-5(x - 10)}{3(x + 6)}; x \neq 9; x \neq 10 $